Chapter 1, Problem 53RE

(a)

To determine

To describe: the graphs in the xy -plane.

Answer to Problem 53RE

The graph of $x=0$ in the xy -plane is a vertical line known as y -axis

Explanation of Solution

Given:

  $x=0$

Calculation:

The graph of $x=0$ in the xy -plane is a vertical line known as y -axis and has undefined slope. It does not have x -intercept.

To graph $x=0$ by hand, look for all the points $\left(x,y\right)$ in the plane for which $x=0$ . No matter what y coordinate is used, the corresponding x coordinate is always equal to zero. Consequently, the graph of $x=0$ in the xy -plane is a vertical line known as y -axis.

Conclusion:

Therefore,the graph of $x=0$ in the xy -plane is a vertical line known as y -axis.

(b)

To determine

To describe: the graphs in the xy -plane.

Answer to Problem 53RE

The graph of $y=0$ in the xy -plane is a horizontal line known as x -axis.

Explanation of Solution

Calculation:

The graph of $y=0$ in the xy -plane is a horizontal line known as x -axis and has undefined slope. It does not have y -intercept.

To graph $y=0$ by hand, look for all the points $\left(x,y\right)$ in the plane for which $y=0$ . No matter what x coordinate is used, the corresponding y coordinate is always equal to zero. Consequently, the graph of $y=0$ in the xy -plane is a horizontal line known as x -axis.

Conclusion:

Therefore, the graph of $y=0$ in the xy -plane is a horizontal line known as x -axis.

(c)

To determine

To describe: the graphs in the xy -plane.

Answer to Problem 53RE

The graph of $x+y=0$ in the xy -plane is a straight line with slope is - 1 and y -intercept is zero.

Explanation of Solution

Calculation:

The graph of $x+y=0$ in the xy -plane is a straight line with negative slope (-1). Its y -intercept is zero.

To graph $x+y=0$ by hand, write the equation in the form of $y=\left\{\text{expression of }x\right\}$

  $\begin{array}{l}x+y=0\\ y=-x\end{array}$

Now, compare to the standard form of equation of line $y=mx+b$ , where m is slope of line and y -intercept is b .

Consequently, the graph of $x+y=0$ in the xy -plane is a straight line with slope is - 1 and y -intercept is zero.

Conclusion:

Therefore, the graph of $x+y=0$ in the xy -plane is a straight line with slope is - 1 and y -intercept is zero.

(d)

To determine

To describe: the graphs in the xy -plane.

Answer to Problem 53RE

The graph of $xy=0$ in the xy -plane is two intersecting lines.

Explanation of Solution

Calculation:

The graph of $xy=0$ in the xy-plane is two lines intersecting each other.

The equation $xy=0$ ; only if $x=0$ ; or $y=0$ or both xand $y=0$ .

Since, graph of $x=0$ is y -axis and graph of $y=0$ is x -axis. Consequently, the graph of $xy=0$ in the xy -plane is two intersecting lines.

Conclusion:

Therefore, the graph of $xy=0$ in the xy -plane is two intersecting lines.

(e)

To determine

To describe: the graphs in the xy -plane.

Answer to Problem 53RE

The graph of $x^2+y^2=0$ in the xy -plane is a point at origin.

Explanation of Solution

Calculation:

The graph of $x^2+y^2=0$ in the xy -plane is a point at origin.

To graph $x^2+y^2=0$ by hand, compare to the standard form of equation of circle

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

So, the center of $x^2+y^2=0$ in the xy -plane is $\left(0,0\right)$ and radius is zero. Consequently the graph of $x^2+y^2=0$ in the xy -plane is a point at origin.

Conclusion:

Therefore, the graph of $x^2+y^2=0$ in the xy -plane is a point at origin.